Helpppp!!!!

• December 2nd 2007, 04:32 PM
d690
Helpppp!!!!
Heres the question exactly as is:
Tangents to the curve f(x)=4x-x^2 intersect at (7/4,11/2). P and Q are points of tangency to the curve. Find the point of intersection of the normals drawn to the curve at points P and Q.

I can only find the point on the left of the curve but i cant find anything about the point ont he right. helppp mee (sorta urgent unit final tomorrow).
(p.s. the answers are (1/2,7/4) its not the anwser im after how to do it)
• December 2nd 2007, 05:24 PM
galactus
You can find the derivative of y to find the slope, m, at the points.

Try using $y-y_{1}=m(x-x_{1})$

$(4x-x^{2})-\frac{11}{2}=(4-2x)(x-\frac{7}{4})$

Solve for x. That will be the x-coordinates of the points of intersection of the two tangent lines.

You can then find their equations and it's downhill.
• December 2nd 2007, 05:50 PM
d690
thanks i got it